If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), t

1.	If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form (A) y – x = 0 (B) x + y = 0 (C) –2x + y = 0 (D) –x + 2y = 0
Answer» (B) x + y = 0 Let us consider a linear equation ax + by + c = 0 … (i) Since, (-2,2), (0, 0) and (2, -2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get At point(-2,2), -2a + 2b + c = 0 …(ii) At point (0, 0), 0+0 + c = 0 => c = 0 …(iii) and at point (2, – 2), 2a - 2b + c = 0 …(iv) From Eqs. (ii) and (iii), c = 0 and – 2a + 2b + 0 = 0, – 2a = -2b,a = 2b/2 => a = b On putting a = b and c = 0 in Eq. (i), bx + by + 0 = 0 => bx + by = 0 => – b(x + y) = 0 => x + y = 0, b ≠ 0 Hence, x + y= 0 is the required form of the linear equation.

If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form (A) y – x = 0 (B) x + y = 0 (C) –2x + y = 0 (D) –x + 2y = 0

Answer»

(B) x + y = 0

Let us consider a linear equation ax + by + c = 0 … (i)

Since, (-2,2), (0, 0) and (2, -2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get

At point(-2,2), -2a + 2b + c = 0 …(ii)

At point (0, 0), 0+0 + c = 0 => c = 0 …(iii)

and at point (2, – 2), 2a - 2b + c = 0 …(iv)

From Eqs. (ii) and (iii),

c = 0 and – 2a + 2b + 0 = 0, – 2a = -2b,a = 2b/2

=> a = b

On putting a = b and c = 0 in Eq. (i),

bx + by + 0 = 0

=> bx + by = 0

=> – b(x + y) = 0

=> x + y = 0, b ≠ 0

Hence, x + y= 0 is the required form of the linear equation.